A hot-water radiator has a surface temperatue of 80 o C and a surface area of 2 m2 . Treating it as a blackbody, find the net ra
1 answer:
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Answer:
Explanation:
From the question we are told that:
Distance of wall from CD
Second bright fringe
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Strontium vapor laser has a wavelength \lambda= 431 nm=>431 *10^{-9}m
Generally the equation for Interference is mathematically given by
Where
Answer:
The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.
Explanation:
As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:
Since the frictional force is equal to
, then we have that the acceleration is:
Now, from the definition of acceleration we get:
And, as the final velocity is zero because the box gets to a stop, we have:
(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)
Then, plugging in the given values, we calculate the time:
In words, the time the box takes to stop sliding relative to the belt is 0.139s.
The displacement of the box in this time, is given by the kinematics formula:
Finally, we calculate the displacement:
This means that the box moves 7.29cm backwards relative to the belt.
Answer:
Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g
Explanation:
The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.
And sum of upward forces = sum of downward forces.
N = mg cos θ
m = 100 g = 0.10 kg
g = acceleration due to gravity = 9.8 m/s²
θ = 15°
N = (0.1×9.8×cos 15°) = 0.946582 N
The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).
For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.
Frictional force = Tension + F
Frictional force = μN
where μ = coefficient of kinetic friction = 0.60
N = normal reaction = 0.9466 N
Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N
The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ
F = (0.10 × 9.8 × sin 15°) = 0.253624 N
Tension in the rope = T = ?
Fr = F + T
T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N
But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.
2T = mg
2 × 0.3144 = 9.8m
m = 0.06416 kg = 64.16 g.
Mass of the container = 30 g
So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g
Hope this Helps!!!
